rhombus tilings: decomposition and space structure
TRANSCRIPT
Laboratoire de l’Informatique du Parallelisme
Ecole Normale Superieure de LyonUnite Mixte de Recherche CNRS-INRIA-ENS LYON no 5668
Rhombus tilings:decomposition and space structure
Frederic ChavanonEric Remila
June 2004
Research Report No 2004-30
Ecole Normale Superieure de Lyon,46 Allee d’Italie, 69364 Lyon Cedex 07, FranceTelephone : +33(0)4.72.72.80.37 Fax : +33(0)4.72.72.80.80 Adresseelectronique :[email protected]
Rhombus tilings:decomposition and space structure
Frederic ChavanonEric Remila
June 2004
Abstract
We study the spaces of rhombus tilings, i.e. the graphs whose verticesare tilings of a fixed zonotope, and two tilings are linked if one can passfrom one to the other one by a local transformation, called flip.We first use a decomposition method to encode rhombus tilings andgive a useful characterization for a sequence of bits to encode a tiling.In codimension 2, we use the previous coding to get a canonical repre-sentation of tilings, and an order structure on the space of tilings, whichis shown to be a graded poset, from which connectivity is deduced.
Keywords: tilings, structure, order
Resume
Nous etudions les espaces de pavages rhombiques, i.e. les graphes dontles sommets sont les pavages d’un zonotope fixe, et deux pavages sontlies si on peut passer de l’un a l’autre par une serie de transformationslocales appelees flips.Nous utilisons une methode de decomposition pour coder ces pavages,et donnons une caracterisation des sequences de bits codant effective-ment des pavages.En codimension 2, nous utilisons ce codage pour donner unerepresentation canonique des pavages, et une structure d’ordre surl’espace des pavages. Cet ordre est gradue, ce qui nous permet d’endeduire la connexite de l’ensemble.
Mots-cles: pavage, structure, ordre
Rhombus tilings:decomposition and space structure
Frederic Chavanon ∗ Eric Remila †
Abstract
We study the spaces of rhombus tilings, i.e. the graphs whose vertices are tilings of afixed zonotope, and two tilings are linked if one can pass from one to the other one by alocal transformation, called flip.
We first use a decomposition method to encode rhombus tilings and give a usefulcharacterization for a sequence of bits to encode a tiling.
In codimension 2, we use the previous coding to get a canonical representation oftilings, and an order structure on the space of tilings, which is shown to be a gradedposet, from which connectivity is deduced.
1 Introduction
Rhombus tilings are tilings of zonotopes with rhombohedra. More precisely, we fix a se-quence (v1, v2, . . . , vD) of vectors of Rd (such that each subsequence (vi1 , vi2 , . . . , vid) oflength d is a basis of Rd) and a sequence (m1, m2, . . . , mD) of positive integers. The tiledzonotope V is the set: V =
{v ∈ Rd, v =
∑Di=1 λivi, 0 ≤ λi ≤ mi,mi ∈ M, vi ∈ V
}, and each
tile is (a translated copy of) a rhombohedron defined by a set {vi1 , vi2 , . . . , vid)} of d vectors.The notion of flip on tilings is carefully studied. Assume that a tiling T of Z contains d + 1tiles which pairwise share a facet. In such a case, a new tiling Tflip of Z can be obtained by aflip consisting in changing the position of the d+1 previous tiles. The space of tilings of Z isthe graph whose vertices are tilings of Z and two tilings are linked by an edge if they differby a single flip. The structure of spaces of tilings in very interesting, since rhombus tilingsappear in physics as a classical model for quasicrystals [1].
In the first part of this paper, we use ideas (deletion, contraction) issued from matroidtheory [2, 8] to get a decomposition method for tilings (section 3). We see how to encodea tiling by a sequence of small tilings, containing d + 1 tiles. Informally, this encoding canbe seen as a pilling of (hyper)cubes, in a similar way as it has been done by Thurston [13]for the particular case of tilings with lozenges (d = 2 and D = 3) for any simply connectedpolygon. The first important result of the paper (the reconstruction theorem, section 4) is a
∗Laboratoire de l’Informatique du Parallelisme, umr 5668 CNRS-INRIA-UCB Lyon 1-ENS Lyon, 46 alleed’Italie, 69364 Lyon cedex 07, France. [email protected]
†Laboratoire de l’Informatique du Parallelisme, umr 5668 CNRS-INRIA-UCB Lyon 1-ENS Lyon, 46 alleed’Italie, 69364 Lyon cedex 07, France and GRIMA, IUT Roanne, 20 avenue de Paris, 42334 Roanne cedex, [email protected]
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characterization of sequences of small tilings which really encode a (big) tiling. This charac-terization is local in the sense that it can be checked using a set of conditions, each of themusing a bounded number of small tilings.
The complexity of the space of tilings is strongly related to the parameter c = D − d,called the codimension. If c = 0, there exists a unique tiling. For c = 1, it can easily be shown(using the pilling of cube interpretation) that the space of tilings can be directed so as to get astructure of distributive lattice. In Section 5, using the reconstruction theorem, we introducea representation of tilings which allows to prove the main result of the paper: for c = 2, thespace of tilings can be directed so as to get a graded poset (with single maximal and minimalelements). This result clearly induces the connectivity of the space, which is a non-trivialresult linked to the general Baues problem [11] on polytopes. As another consequence, weobtain that flips induce a Markov chain on the space of tilings, whose limit distribution isuniform.
Before this paper, a parallel study has been done by Ziegler [15], about higher Bruhat or-ders. Those are combinatorial structures, which can be interpreted as extensions of matroids.We recall that the Bohne-Dress theorem [12] claims that tilings of zonotopes can also be in-terpreted as extensions of matroids. Thus, the study from Ziegler can be seen as the studyof tilings of unitary zonotopes constructed from vectors in cyclic arrangement. For thosezonotopes, Ziegler proves that the space of tilings can be directed so as to get a graded poset(with a unique minimal element and a unique maximal element), for c ≤ 4. Later, Felsnerand Weil [7] prove the same result, when d ≤ 2. To our knowledge, the connectivity problemis still open for the other kinds of zonotopes. We mention that R. Kenyon [10] has provedthe connectivity in dimension 2, for any simply connected domain.
2 Tilings of Zonotopes and Minors
We deal in this paper with a particular case of tilings in Rd, called zonotopal rhombic tilings.Let us now define the fundamental elements studied in the following.
The canonical basis of Rd will be noticed (e1, e2, ..., ed). Let V = (v1, ..., vD) be a sequenceof D vectors in Rd such that D ≥ d and each subsequence (vi1 , vi2 , . . . , vid) is a basis of Rd.The parameter c = D − d is called the codimension.
Let M = (m1, ...,mD) be a sequence of D nonnegative integers. mi is associated with thevector vi and called the multiplicity of vi. The zonotope Z = (V, M) associated with V and M
is the region of Rd defined by:{
v ∈ Rd, v =∑D
i=1 λivi, 0 ≤ λi ≤ mi,mi ∈ M, vi ∈ V}
. Thus,
Z is the convex hull of the finite set{
v ∈ Rd, v =∑D
i=1 λimivi, λi ∈ {0, 1}, vi ∈ V}
.One can define classically (see for example [14] p. 51-52) its faces, vertices (faces of di-
mension 0), edges (faces of dimension 1), and facets (faces of dimension d− 1). The number:s =
∑Di=1 mi is the size of the zonotope Z; we say that Z is an s-zonotope. Z is said to be
unitary if all the multiplicities are equal to 1 (see Figure 1 for examples).Let Z = (V,M) be a zonotope. The sequence of vectors forming the sequence V is called
the type of Z. A prototile is a unitary zonotope constructed with a subsequence V ′ of d distinctvectors taken in V . A sequence V of D vectors of Rd induces (D
d ) different prototiles. A tilet is a translated prototile, i.e. it is defined by a pair (p, w), where p is a prototile and w atranslation vector (formally, we have: t = w + p). Since tiles are some particular polytopes,their vertices, edges and facets are defined as well. The type of a tile is the type of the
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Figure 1: A 2-dimensional zonotope and a 3-dimensional zonotope both defined on 4 vectors
corresponding prototile.A tiling T of a zonotope Z = (V,M) is a set of tiles constructed with vectors in V , such
that each intersection between tiles is a face of the tiles (i.e. there is no overlap) and the unionof tiles is equal to Z (i.e. there is no gap). The extremal points of a tile are the vertices of thetiling, and the edges of a tile are the edges of the tiling. Two tiles are adjacent if they share awhole facet. We say that Z is the support of the tiling T . If Z is an s-zonotope, we say that Tis an s-tiling.
2.1 De Bruijn sections
Definition 1 (lifting, height function) Let Z = (V, M) be a zonotope, with V = (v1, ..., vD),and x0 be a fixed extremal point of Z.
A lifting of V is a sequence U = (u1, ..., uD) of vectors of Rd+1, for each integer j such that1 ≤ j ≤ D, we have: uj = (vj , αj).
Let T be a tiling of Z. The associated lifting is the unique function fT,U which associates toeach vertex of TZ a vector in Rd+1 and satisfies the following properties: fT,U (x0) = (0, 0, . . . , 0)and for any pair (x, x′) of vertices of T such that x′ = x + vi and [x, x′] is an edge of T , we have:fT,U (x′) = fT,U (x) + ui. See Figure 2.
The height function hT,U associated with a lifted tiling fT,U , is the component upon ed+1 offT,U .
One can easily prove that the definition of lifting of a tiling is consistent since a zonotopeis homeomorphic to a closed disk of Rd. Notice that fT,U is defined for the set of vertices ofZ and does not depend on the tiling T chosen.
The two mostly used lifting functions are the principal lifting function, defined by ∀vi ∈ V ,ui = (vi, 1), and the k-located function, where for a fixed integer k, uk = (vk, 1) and ∀i 6= k,ui = (vi, 0) . The k-located function has same value on each vertex of a tile whose type doesnot contain vk, and differs by 1 at the endpoints of an edge of type vk. Consequently, theprincipal function differs by 1 at the endpoints of each edge of the tiling.
Now, since height functions have been defined, one may introduce the important conceptof de Bruijn families and sections, widely used in the core of the paper (See [5] for details).
Definition 2 (de Bruijn section, family) Let T be a tiling of a zonotope Z = (V, M), and hi thei-located function. The de Bruijn family associated with the vector vi is the set of tiles having vi intheir type. Moreover, the j-th de Bruijn section is the set of tiles whose i-located function is j − 1on one facet, and j on the opposite facet. This section will be noted S{vi,j} (see Figure 2).
3
2
1
1
1
11
1
1
0
0
0
0
0 0 0
{v ,1}
{v ,2}
2
2
2
2
2
2
2
2
2 2 2
2
2
Figure 2: The 2-located height function and two de Bruijn sections.
One sees that a de Bruijn section S{vi,j} disconnects the tiling into two parts, T+{vi,j} and
T−{vi,j}. The first one is composed of tiles whose vertices have i-located function larger thanj, and the second corresponds to the tiles whose height function is smaller. Hence, for j < j′,we have: T−{vi,j} ⊆ T−{vi,j′}.
We say that two de Bruijn sections S{vi,j} and S{vk,l} are parallel if vi = vk. The intersec-tion of a set of d de Bruijn sections of T which are pairwise not parallel is a tile of T . Theintersection of a set of d− 1 de Bruijn sections which are pairwise not parallel is a set of tileswhich can be totally ordered in such a way that two consecutive tiles are adjacent. Such anintersection is called a de Bruijn line.
2.2 Flips
2.2.1 Tilings of a unitary d+1-zonotope
We first focus on a unitary zonotope of codimension 1. One easily checks that such a zono-tope admits exactly two tilings: Let V = (v1, ..., vd+1) be the sequence of vectors and p0 be theprototile constructed with the d first vectors: there is a tiling T with a tile t0 of type p0 suchthat T+
{vd+1,1} is empty and T−{vd+1,1} = {t0}, and one tiling T ′ such that T ′+{vd+1,1} = {t0 +vd+1}and T ′−{vd+1,1} is empty.
One arbitrarily considers one of these tilings as the high position, and the other one as thelow position.
Remark that T and T ′ are symmetrical according to the symmetry centered in the vector:∑1≤i≤d+1 vi/2.Any pair of tiles of T (or T ′) are adjacent, since they form a whole de Bruijn line of T . The
orders in each de Bruijn line are opposite in T and T ′.
2.2.2 Space of tilings
Those tilings of unitary zonotope of codimension 1 can appear, translated, in tilings of alarger zonotope Z. Assume that the tiling Tz of a unitary d+1-zonotope z of codimension 1appears in a tiling T of Z, translated by a vector v (i. e. formally: v + Tz ⊂ T ). We say that
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the tiling T ′ of Z, defined by: T ′ = (T \ (v + Tz))∪ (v +T ′z), is obtained from T by a geometricflip.
The type of the flip is the type of z. It will be denoted by the set of indexes of vectors ofits support. We have (D
d+1) types of flips; in particular, for D = d + 2, we have D types offlips. The flip is an up flip if Tz is the low tiling of z, and its converse is a down flip.
The space of tilings of a zonotope Z is the symmetric labeled graph whose vertices are thetilings of Z, and two tilings are linked by an edge if they differ by a geometric flip. The labelof the edge is the type of the corresponding flip.
The edges of the space of tilings can then be directed by the orientation of flips. Onecan set a relation between tilings as follows: given two distinct tilings T and T ′, we say thatT <flip T ′ if there exists a sequence of up flips leading from T to T ′. (i.e. if there exists adirected path from T to T ′ in the space of tilings, directed as above). We will see later that<flip is really an order relation, which is not a priori true, with this definition.
In the following, we label a flip by the indexes of the vectors defining the flipped tiles.An important result is that flips induce connectivity between all tilings of zonotopes for
d = 2, i.e. every tiling of a given dimension 2 zonotope Z can be deduced from another tilingof Z by a sequence of flips (see [4, 6, 10] for details). This is an important open question inthe case of larger dimensions.
The point is now to study spaces of zonotopal tilings. Despite the fact that rhombic tilingsare defined for any dimension, the figures are in dimension 2, for convenience.
2.3 Deletions
Let T be a tiling of support T , and S{vi,j} be a de Bruijn section of T . One can remove the tilesof S{vi,j} and translate all the tiles of T+
{vi,j} by the vector −vi. For D > d, the configurationobtained is a tiling of Z ′ = (V,M ′) where M ′ is defined by: m′
i = mi−1 and ∀k 6= i,m′k = mk
(except in the special case when mi = 1, in such a case, we have Z ′ = (V ′,M ′) with V ′ andM ′ respectively obtained from V and M removing the ith component). Such an operationdefines a deletion relation on zonotope tilings.
The tiling obtained is denoted by D{vi,j}(T ), and for each tile t of T , we state: D{vi,j}(t) =t for t in T+
{vi,j}, and D{vi,j}(t) = t− vi for t in T+{vi,j}.
For consistence, the de Bruijn sections of D{vi,j}(T ) according to vi are assumed to benumbered 1, 2, . . . , j − 1, j + 1, . . . ,mi. By this way, D{vi,j}(t) and t both are in de Bruijnsections with the same label. We also need this convention for the proposition below, whenvi = vk.
Proposition 2.1 (commutativity of deletions) Let T be a tiling of a zonotope Z , and two dele-tions D{vi,j} and D{vk,l}. We have:
D{vi,j}(D{vk,l}(T )) = D{vk,l}(D{vi,j}(T ))
Proof: The tiling T can be partitioned into the five parts below:
• T−{vi,j} ∩ T−{vk,l}: the tiles of this part remain unchanged by the successive deletions,taken in any order,
• T+{vi,j} ∩ T−{vk,l}: the tiles of this part are translated by −vi during the successive dele-
tions, taken in any order,
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• T−{vi,j} ∩ T+{vk,l}: the tiles of this part are translated by −vk during the successive dele-
tions, taken in any order,
• T+{vi,j} ∩ T+
{vk,l}: the tiles of this part are translated by −(vi + vk) during the successivedeletions, taken in any order.
• S{vi,j}∪S{vk,l}: the tiles of this part are removed during the successive deletions, takenin any order.
Thus the order of deletions does not give any change. This gives the commutativity re-sult. ¤
A tiling obtained from T by a sequence of p deletions is called a s−p-minor of T .The pairs {vi, j} can be totally ordered (by the lexicographic order, for example). From
this order, the sets {{vi1 , j1}, {vi2 , j2}, . . . , {vip , jp}} formed by p elements of the type {vi, j}can also be totally ordered. Therefore, the s−p-minors of T can be totally ordered. Thesequence of s−p-minors of T is given by this order.
Figure 3: Commutativity of deletions.
Proposition 2.2 Assuming s ≥ d + 2, every tiling is defined by the sequence of its s−1-minors.
Proof: Let Z be a zonotope, T one of its tilings. Notice that one can easily compute thesequence of s−2-minors of T from the sequence of its s−1-minors. Let {vi, j} and {vi′ , j
′}be two distinct pairs, D1 and D2 respectively denote the corresponding deletions, and D1,2
denote the corresponding double deletion.For each tile t′ of D1,2(T ), one can easily compute the tiles t1 such that t1 is in D2(T ) and
D1(t1) = t′, and t2 such that t2 is in D1(T ) and D2(t2) = t′. Precisely, on can compute thepair (ε1, ε2) of {0, 1}2 such that t1 = t′ + ε1vi and t2 = t′ + ε2vi′ .
Let t0 be the tile of T such that t1 = D2(t0). From the commutativity, we also have:t2 = D1(t0) (see Figure 4). Moreover, from the definition of de Bruijn sections, t0 = t2+ε1vi =t′ + ε1vi + ε2vi′ : the tile t0 can be computed from t′, t1 and t2.
This gives the result, since for each tile t of T , there are two distinct pairs {vi, j} and{vi′ , j
′} such that t is out of S{vi,j} ∪ S{vi′ ,j′} (from the hypothesis: s ≥ d + 2). ¤
Notice that the result is false for: s = d + 1. Each d-minor is reduced to a single tile, thusthe information about the arrangement of tiles is lost.
6
1
0t
12 tt
T
1,2D
D
D
D
D
t’
2
1
2
Figure 4: Proof of Proposition 2.2: computation of some tiles of T from D2(T ) and D1(T )
Iterating the proof for (s− 1)-deletion, one obtains the following result as a corollary forproposition 2.2 (see Figure 3).
Corollary 2.3 Let s′ be a integer such that d + 1 ≤ s′ ≤ s. Assuming s ≥ d + 2, every tiling T ofzonotope is defined by the sequence of its s′-minors.
In particular, this is true for d+1-minors.
Proof: Obvious by induction. ¤
Remark that there are two kinds of d+1-minors: those of codimension 0, the forced minors,which are defined by the tiled zonotopes, and those of codimension 1, the free minors, whichare tilings of unitary zonotopes. Only the free ones contain some information, useful tocompute T . This information can be reduced to a single bit, corresponding to the fact that thezonotope is respectively in low or high position. This gives an encoding of zonotope tilingsby a word on the alphabet {0, 1} of length
∑1≤i1<i2<...<id≤D mi1mi2 . . .mid (see Figure 5 for
an example).We define a set flip as follows: let T and T ′ be two tilings of a same zonotope such that
all their d+1-minors are the same, except one. we say that T and T ′ differ by a set flip. Wesay that the set flip from T to T ′ is an up flip if the different d+1-minor is in low position inT (and in high position in T ′).
Proposition 2.4 let T and T ′ be two tilings of a zonotope Z. T differs from T ′ by an upward set flipif and only if T differs from T ′ by an upward geometric flip.
The relation <flip defines a partial order.
Proof: It is clear that a geometric flip is a set flip with the same orientation, because itchanges locally the positions of d + 1 tiles. Since only one d+1-minor contains all these tiles,their positions are changed in only this minor.
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Low positions
0 0 1 1
D{v1}=234
1
D{v2}=134
0
D{v3}=124
1
D{v4}=123
0
Figure 5: Coding of tilings with d+1-minors.
For the converse part, we first study how a deletion and a set flip act on a fixed de Bruijnline dBL. A deletion (which does not remove the whole de Bruijn line dBL) only removesone tile of dBL and does not change the order in this line for the other tiles. Thus a set flipchanges the order on dBL if and only if dBL contains a pair {t, t′} of tiles which appear inthe flip. Moreover, in this case, the comparison order is changed only for the pair {t, t′}, sinceany other pair of tiles appears in a d+1-minor unchanged by the flip. Thus, for consistence ofthe order, the tiles t and t′ necessarily share a whole facet. Thus the flip is actually geometric.
The second part of the Proposition is an obvious corollary of the first part. ¤
3 Reconstruction theorem
We are interested in the following problem: given a zonotope Z and a sequence of d+1-tilings (with the good length, and the good vectors), does there exist a tiling T of Z such thatthe given sequence is the sequence of its d+1-minors ?
We can obviously solve the problem by constructing the (potential) d+2-minors, thenthe d+3-minors, and so on until the searched tiling is found. If there is a contradiction, thereconstruction is impossible, otherwise the tiling is obtained. But this can give the answerfaster, and the following proposition states that the first step is enough to obtain the answer.
Proposition 3.1 Let (Zm)m be a sequence of d+1-tilings. There exists a tiling T of a zonotope Zwhose sequence of d+1-minors is exactly (Zm)m if and only if the d+2-minors are compatible, i.e.the sequence of d+2-minors can be correctly constructed.
Proof: We do the proof by induction on the size s of the zonotope. The case s = d + 2 isobvious.
Let s > d + 2. Consider the prefix of the sequence (Zm)m formed by d+1-tilings wherethe tiles of the (potential) de Bruijn section {vD,mD} do not appear (since it is assumed thatthe deletion D{vD,mD} has been done). This subsequence is, by assumption, the sequence ofd+1-minors of a tiling T ′ of size s− 1.
On the other hand, for each tile t of T ′, there exists a d+1-tiling Tt containing t and d tilesof the de Bruijn section {vD,mD}. Hence t can be assigned a + or − sign, depending on its
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position in Tt, relatively to S{vD,mD} (+ if the D-located height function of t is 1, − if it is0). Let T ′+ be the part of T ′ formed by the tiles marked + and T ′− the part formed by tilesmarked −.
Let us now consider a straight line l directed by vD. We claim that, following l in thesense of vD, one first meets tiles marked −, then tiles marked +. This means that T ′+ andT ′− are convex along vD, i.e. that the new de Bruijn section can be inserted correctly in T ′,thus leading to a new tiling T . Two cases may occur:
• l only meets facets and interior parts tiles of T ′. Consider two tiles of T ′, say t1 and t2,which share a facet, and such that t2 follows t1 in the succession of tiles crossed by l inthe direction of vD. There exists a d+2-minor Td+2 containing (tiles corresponding to)tiles of {vD,mD} and tiles t1 and t2. There are only three possible sign assignment for(t1, t2), since the assignment: + for t1 and − for t2, is impossible ; otherwise the tile t3of type {τ} ∪ {vD} (where τ denotes the set of common vectors in the types of t1 andt2) cannot be placed in the d+2-minor Td+2 (see Figure 6).
T
t
1tt1
t2
t2
t2
1t
t2
−−
−
+++
Dv
−
+T
1
Figure 6: The three possible sign assignments for t1 and t2, from the possible positions in thede Bruijn line of Td+2.
• l meets a face f of the tiling T ′ of dimension lower than d− 1. Then, there are two tilest1 and t2 with the same hypothesis as in the previous case, but sharing only the facef . There exists a parallel line l′, arbitrarily close to l, satisfying the hypothesis of theprevious case, and crossing both t1 and t2 (but t1 and t2 are not necessarily consecutivealong l′. See Figure 7). Thus the assignment: + for t1 and − for t2, is impossible.
l
21t
t
l’
Figure 7: A line l crossing a vertex, and the auxiliary line l′
.
Hence T ′+ and T ′− are consistent according to {vD,mD}, allowing to translate the part
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T ′+ by vD, in order to insert the de Bruijn section S{vD,mD}. The tiling T obtained (such thatT ′+ = T+
{vD,mD} + vD and T ′− = T−{vD,mD}) is the one searched, which ends the proof. ¤
4 Representation in codimension 2
From this section, we limit ourselves to the case when D = d + 2. A type of a flip will belabeled by the sequence of indexes of the d+1 vectors involving in it, or, for convenience, bythe overlined index by the missing vector (for example, a flip whose support is the unitaryzonotope constructed with the first d + 1 vectors is labeled by 12...d + 1 = d + 2).
As seen formerly, the zonotopal tilings can be easily encoded by considering their minors.More precisely, one tiling is defined by the set of high or low positions of all (free) d+1-minors. We will now describe a representation tool for zonotope tilings based on the minorstructure. But, before doing it, we need more knowledge about d+2-zonotopes.
4.1 The basic d+2-zonotopes
In dimension d, there exists two basic kinds of d+2-zonotopes of dimension d whose tilingis not forced: either all vectors have multiplicity 1 (codimension 2), or there is one vector ofmultiplicity 2 (codimension 1). We first precisely study these cases.
4.1.1 The d+2-zonotope of codimension 1
Proposition 4.1 The space of tilings of the zonotope Zi of codimension 1 with the vector vi of multi-plicity 2 (and the d other ones of multiplicity 1) contains three tilings and is a chain of length 2.
Proof: In each tiling, there exists a unique tile t which is not element of S{vi,1} ∪ S{vi,2}.Since T−{vi,1} ⊆ T−{vi,2}, we have three tilings:
• one tiling with t ∈ T−{vi,1},
• one tiling with t ∈ T+{vi,2},
• one tiling with t ∈ T−{vi,2} \ T−{vi,1}.
The directed edges corresponding to flips are obvious (remark that both the free d+1-minors of Zi are of the same type, which gives the chain) (see Figure 8). ¤
4.1.2 The unitary d+2-zonotope
We first need more information about the structure of tilings of unitary d+1-zonotopes. Thisis given by the lemma below.
Lemma 4.2 Let T be a tiling of unitary d+1-zonotope, and v be a vector not in the type of Z. Wedefine a tournament G(T,v) on the tiles of T saying that (t1, t2) is an arc of G(T,v) if the vector vcrosses their common facet passing from t1 to t2 (see Figure 9).
The tournament G(T,v) is actually a total order.
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Figure 8: Space of tilings of a codimension 1 zonotope with one duplicated vector.
31
2
Figure 9: A codimension 1 tiling, the added vector (dashed), and the ordering of tiles accord-ing to this vector.
Proof: Since all pairs of tiles are linked, we only have to prove that G(T,v) has no cycle oflength 3. We prove it reducing the problem to the case: d = 2, for which the proof is easy bya case by case analysis.
In higher dimension, notice that, since D = d + 1, the types of three given tiles t1, t2 andt3 contain (d + 1) − 3 = d − 2 common vectors. Let p denote the orthogonal projection onthe 2-dimensional space which is orthogonal to the d − 2 common vectors. The projectionsp(ti) are parallelograms, and we have (ti, tj) in G(T,v) if and only if (p(ti), p(tj)) is in GH,p(v);where H denotes the hexagon covered by the parallelograms p(ti) (H is really a hexagon,since otherwise the tiles ti cannot be pairwise adjacent). This gives the result, since Gp(v) isnot a cycle. ¤
Proposition 4.3 The space of tilings of a unitary d+2-zonotope is a cycle of length 2(d + 2), andeach possible label is given to a pair of edges, which are opposite in the cycle.
Proof: let T be a tiling of the unitary zonotope Z = ((v1, v2, . . . , vd+2), (1, 1, . . . , 1)). Fromthe above lemma applied on the support Z ′ of D{vd+2,1}(T ), T−{vd+2,1} is an initial segment ofthe order induced by vd+2 on tiles of D{vd+2,1}(T ).
Conversely, given a tiling T ′ whose support is Z ′, and an initial segment T ′ (accordingto the order induced by vd+2), one easily constructs a tiling of Z: tiles which are not in theinitial segment are translated by vd+2, and tiles of S{vd+2,1} are inserted in the remainingspace. There exists d + 2 possible initial segments for a set of d + 1 elements, thus, since Z ′
admits two tilings, there exists 2(d + 2) tilings of Z.Now, take a tiling of T , i. e. a tiling T ′ of the zonotope Z ′ and I one initial segment of
it. What are the possible flips from T ? First assume that the initial segment is proper (i. e.
11
neither empty nor equal to T ′). There are two possible flips, which correspond to adding orremoving one tile in I . No other flip is possible because of the relative position of tiles givenby the order on tiles of T ′ (the flip only using tiles of T ′ is not possible because of the cut byS{vd+2,1}).
A similar argument holds for the other case. If I is empty, two flips are possible, onewhich corresponds to adding the first tile in I , the other one only uses tiles of T ′. If I = T ′,two flips are possible, one which corresponds to removing the last tile in I , the other oneonly uses tiles of T ′. This gives the result, using the symmetry of both tilings of Z ′ to get thelabels of opposite edges. ¤
4.1.3 Order convention
We fix a basic tiling T0 of the unitary d+2-zonotope. We enforce the convention of lowand high position, defining low positions of d+1-tilings as tilings which are elements of thesequence of d+1-minors of T0. With this convention, T0 is the lowest tiling of the unitaryd+2-zonotope; T1, the opposite tiling in the cycle, is the largest tiling, The space is formedby two directed chains from T0 to T1, which only meet at their endpoints. Up to vectorrenumbering , it can be assumed that, from T0 to T1 the sequence of successive labels ina chain is (d + 2, d + 1, . . . , 1) and the sequence is (1, 2, . . . , d + 2) for the other chain (seeFigure 10).
We can encode each tiling by its position on the cycle as follows: the tiling at the end ofthe path issued from T0, whose sequence of labels is (d + 2, d + 1, . . . , d + 3− i), is denotedby TiL (informally, TiL is in ith position on the left chain), and the tiling at the end of thepath, issued from T0, whose sequence of labels is (1, 2, . . . , i), is denoted by TiR (informally,TiR is in ith position on the right chain).
T
3R
2L2R
1L 1R123
123
124
124134
134
234
2343L
T
T T
TT
Figure 10: The order associated with a unitary octagon.
4.2 Tiling diagrams
We can now precisely explain how we represent a fixed tiling T .
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4.2.1 Points
In our representation, each d+1-minor is associated to a point. Each point p is defined by twoparameters. We first have a coordinate vector, element of Zd+2, which indicates the position ofthe d+1-minor in the sequence of minors: The ith component, denoted by ip, of this vector isequal to j if the deletion according to the pair {vi, j} has not been done to obtain the corre-sponding d+1-minor; the component ip is null if, for each integer j such that 1 ≤ j ≤ mi, thedeletion according to {vi, j} have been done (thus there exists exactly one null component).Remark that a similar vector coordinate will also be given to each d+2-minor whose supportis a unitary zonotope: the only difference is that there is no null component.
The other parameter is a color, which is white if the d+1-minor is in low position, or blackif in high position.
The important thing for reconstructing a tiling T is the set of coloring constraints whichare given by the sequence of d+2-minors. We now explain how coloring constraints areexpressed.
4.3 Arrows
Two points correspond to the pair of minors of a same d+2-minor (the support of this d+2-minor is a d+2-zonotope of codimension 1) if and only if they only differ by one non-nullcoordinate.
From what has been seen about these d+2-tilings, there exists exactly three allowed col-orings of such a pair of points, corresponding to tilings of a d+2-zonotope.
The forbidden coloring uses both colors. We put an arrow in such a way that the originof the arrow is black in the forbidden coloring (and the end of the arrow is black). Thus, thethree allowed colorings of the tiled d+2-zonotope are the fully black one, the fully white one,and the coloring with the origin of the arrow being black and the other point being white.This is the first constraint.
The arrow is labeled by the index of the coordinate which is different for the points.The arrows of the diagram give the covering relation: a point p is covered by a point p′ if
there exists an arrow such that p is the origin of the arrow, and p′ its endpoint.
4.4 Lines
Fix a d+2-minor whose support is a unitary d+2-zonotope. A point corresponds to a d+1-minor of this d+2-tiling if and only if its coordinate vector is obtained replacing one coordi-nate of the d+2-minor by 0.
Such points are linked by line, i. e. they form a sequence ordered in the same way as flipsare ordered in a path between T0 and T1. From what has been seen about tilings of unitaryd+2-zonotopes, with the order convention, the black points have to form a final or initialsegment (i. e. a suffix or a prefix) of the line. This is the second constraint.
Hence, tilings of zonotopes are presented as diagrams on which lines represent unitaryd+2-zonotopes, and arrows represent d+2-zonotopes of codimension 1 (see Figure 11). No-tice that arrows and lines only depend on the support of the tiling. From the reconstructiontheorem, a coloring of points induces a tiling if and only if it respects the constraints above.
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(1,2,1,0)
(1,1,0,1)
(1,0,1,1)
(0,2,1,1)
(0,1,1,1)
l : (1,1,1,1)
l’ : (1,2,1,1)
{v2,1}
{v1,1}
{v4,1}
{v3,1}
{v2,2} (1,2,0,1)
(1,1,1,0)
Figure 11: A tiling and the associated diagram (notice the orientation of the arrows, accord-ing to the inversion property).
4.5 Properties of diagrams
First notice that the highest diagram (i. e. with all points black) and the lowest one (i. e. withall points white) are tilings. We will use two main properties of diagrams.
Lemma 4.4 (inversion property) Let l = (p1, p2, . . . , pd+2) and l′ = (p′1, p′2, . . . , p′d+2) be two
distinct lines such that there exists a unique integer i such that pi = p′i.Assume p1 is covered by p′1. For any integer j such that 1 ≤ j < i, pj is covered by p′j , and for
any integer j such that i < j ≤ d + 2, p′j is covered by pj (see Figure 11 for an illustration of thisproperty).
Of course, a similar property holds when it is assumed that pd+2 is covered by p′d+2.
Proof: Consider the d+2-minor whose sequence of minors corresponds to points pi and p′i.Its support Z0 has codimension 2, its multiplicity is 1 according to any vector vj , for j 6= iand its multiplicity is 2 according to the vector vi.
Consider the tiling Twh. of Z0 corresponding to the fully white coloring. This tiling hastwo minors of codimension 2 which are obtained by a deletion according to vi. By definition,both these minors are equal to T0. That means that there is no tile between both de Bruijnlines according to vi of Twh..
By a sequence of flips, one can move the de Bruijn lines in such a way that each tile(whose type does not contain vi) of the resulting tiling Tinside is between both de Bruijnlines according to vi (see Figure 12). This is done using two successive sequences of flips,one labeled by (1, 2, . . . , i− 1) used to move a de Bruijn line, and the other one labeled by(d + 2, d + 1, . . . , i + 1) to move the other de Bruijn line.
Now, the set of black points is one of the sets: {p1, p2, . . . , pi−1, p′i+1, p′i+2, . . . , p′d+2} or{p′1, p′2, . . . , p′i−1, pi+1, pi+2, . . . , pd+2}. But the second set is not possible, from our assump-tion about the arrow from p1 to p′1 (if p′1 is black, then p1 is necessarily black).
Thus the set of black points is {p1, p2, . . . , pi−1, p′d+2, p′d+1, . . . , p′i+1}, which forces the
sense of arrows, and gives the result. ¤
The inversion property allows to define the labeled line graph whose vertices are the linesof the diagram. Let l = (p1, p2, . . . , pd+2) and l′ = (p′1, p
′2, . . . , p′d+2) be two distinct lines. The
pair (l, l′) is an edge of the line graph if and only if that there exists a unique integer i such
14
wh TT inside
Figure 12: The tilings Twh and Tinside.
that pi = p′i. This edge is labeled by i− if there is an arrow of the diagram from p1 to p′1, orthere is an arrow from p′d+2 to pd+2; otherwise the edge is labeled by i+.
l’1
d+2p’
pd+2p
p2
p’1
2l’l
l’
l
−1l l’
p’
l
pl’
l
+3l l’
−
Figure 13: Examples of edges of the line graph.
The line graph only depends on the support of the tiling, i. e. two tilings with the samesupport induce the same line global line graph (informally, the sign included in the labelindicates the sense of rotation to pass from l to l′).
Lemma 4.5 (consistence property) Let (l1, l′1) and (l2, l′2) be two edges of the line graph whoselabels are elements of the set {i+, i−}. Both these edges have the same label if and only if: (il1 −il′1)(il2 − il′2) > 0.
Proof: Consider the zonotope Z0 of codimension 2, with multiplicity 1 for any vector vj
such that j 6= i and multiplicity 2 for vi. As for the previous lemma, each pair (lk, l′k) inducesa tiling of Z0.
Consider the tiling Twh. of Z0 corresponding to the fully white coloring. Each de Bruijnsection of Twh. according to vi is numbered by an element of the set {ilk , il′k}, according to thepair (lk, lk′). Moreover, the lowest index of each pair {ilk , il′k} is given to the same de Bruijnsection (and the largest index given to the other de Bruijn section).
From this tiling, as in the previous lemma, one can do a sequence of flips in order toobtain the tiling Tinside. Looking at the set of black points for each pair of lines, the result isobtained, considering the set of labels of flips in each case. ¤
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5 Structure of the order in codimension 2
We will now use our representation to obtain some structural results on the space of tilings.Our main theorem is stated below:
Theorem 5.1 Let (T, T ′) be a pair of tilings with the same support, and BT (respectively BT ′) be theset of black points for T (respectively T ′) of the diagram.
We have: T ≤ T ′ if and only if: BT ⊆ BT ′ .
The direct part of the theorem is obvious. To prove the converse, we consider two tilingsT and T ′ such that BT ⊆ BT ′ . The proof uses two lemmas which will be detailed first.
For convenience, we first introduce some vocabulary. The points in BT are said whollyblack, those which are not elements of B′
T ′ are wholly white, the remaining points are positive.A positive point is critical if it is only covered by fully white points. A positive point p isremovable for the pair (T, T ′) if BT ′ \ {p} is still the set of black points of a diagram of tiling.A removable point is necessarily critical. We have to prove that (when T 6= T ′) there exists aremovable point.
If we only consider a tiling, then the lowest one is, by default, the fully white tiling. Wespeak of absolutely removable point.
5.1 Cluster reduction
The cluster Fp generated by a point p is the set of lines passing through p. From Lemma 4.5,the covering relation induces a total order over the points of Fp having same type. Moreover,a cluster can be seen as the diagram of a zonotope Zp (with all multiplicities expect one beingunitary), and the previous colorings, restricted to Fp, are tilings, say Tp and T ′p, of Zp. SinceBT ⊆ BT ′ , we have: BTp ⊆ BT ′p .
The point p is covered by no other point of Fp. Thus, if p is a critical point for (T, T ′), thenp becomes removable in Fp, for the pair (Tp, T
′p).
Lemma 5.2 Let p be a point removable in Fp for the pair (Tp, T′p), and p′ be a positive point which
covers p. The point p′ is removable in Fp′ for the pair (Tp′ , T′p′).
Proof: Consider a line l′ passing through p′. There exists a line l passing through pand meeting l′ (since p′ covers p they differ by only one coordinate) on a common point p′′.Let (Tl, T
′l ) denote the pair of tilings of the unitary d+2-zonotope induced by (T, T ′) from
the colorings of the line l, From the inversion property, one sees (by an easy case by caseanalysis) that if p is removable on l for the pair (Tl, T
′l ), then p′ is removable on l′ for the pair
(Tl′ , T′l′) (see Figure 14 for an example; one can check that other configurations lead to the
same result). Thus p′ is removable in Fp′ . ¤From this lemma, it suffices to prove that there exists a point p which can be removed
in Fp, to prove Theorem 5.1. We will do it now, using an appropriate notion: the obstaclegraph.
5.2 Obstacle graph
We now present the main tool used for the proof of Theorem 5.1.
16
l
p
l’p’
Figure 14: An example for the proof of Lemma 5.2: if p is removable for l, then p′ is removable for l′.
Definition 3 (Obstacle graph) The obstacle graph is the labeled directed graph G where:
• the vertices of G are the tilings (or corresponding lines of colored points) of the unitary d+2-zonotope (except T0),
• (T, T ′) is an edge of G if there exists a d+3-tiling Taux of codimension 2 such that:
– T is a minor of Taux corresponding to a line l = (p1, p2, . . . , pd+2) included in the dia-gram of Taux,
– T ′ is a minor of Taux corresponding to the other line l′ = (p′1, p′2, . . . , p
′d+2) included in
the diagram of Taux, ,– the unique point pi such that pi = p′i is (absolutely) removable in l, but is not (absolutely)
removable in l′.
• If p1 = p′1, then the edge (T, T ′) is labeled by 1+. If the integer i such that pi = p′i is not equalto 1, then the edge (T, T ′) is labeled by i− if p1 is covered by p′1, and labeled by i+ otherwise.
The only edges of the obstacle graph are those listed below (see Figure 15):
• for each pair (k, j) such that 1 ≤ k < j ≤ d + 2, the pair (TkL, TjL) is an edge labeledby k− (except the pair (T1L, T(d+2)L) which is not an edge),
• for each pair (k, j) such that 1 < k < d + 2 and d + 3 < j + k, the pair (TkL, TjR) is anedge labeled by k−,
• for each pair (k, j) such that 1 ≤ k < j ≤ d + 2, the pair (TkR, TjR) is an edge labeledby k+ (except the pair (T1R, T(d+2)R) which is not an edge),
• for each pair (k, j) such that 1 < k < d + 2 and d + 3 < j + k, the pair (TkR, TjL) is anedge labeled by k+,
• for each integer j such that 1 ≤ j < d + 2, the pair (T1, TjL) (we recall T1 = T(d+2)L =T(d+2)R) is an edge labeled by 1+, and the pair (T1, TjR) is an edge labeled by d + 2−,
Lemma 5.3 We say that a directed cycle in the obstacle graph is equilibrated if the sequence of labelsof its edges is such that, for each integer i such that 1 ≤ i ≤ d+2, the label i+ appears in the sequenceif and only if i− also appears.
The obstacle graph has no (non-empty) equilibrated cycle.
17
1R
+2
(n−1)__
_
_
_
_
_
_
+
++
+
+
+
+
(n−1)
(n−1)
2(n−1)n
n
n
n
2
2
11
1
1
n
(n−1)L (n−1)R
3R3L
2L 2R
1L
Figure 15: The obstacle graph.
Proof: Assume we try to find an equilibrated cycle. The vertices T1L and T1R cannot bein the cycle, since they have no incoming edge. Then we remove them and all the edgesadjacent to them, i.e. all the edges labeled by 1− and (d + 2)+. Because of the equilibrationrequirement for the cycle, we remove all the edges labeled by 1+ or (d + 2)−.
We start again with vertices T2L and T2R. Since edges labeled by 1+ and (d + 2)− havebeen removed, T2L and T2R have no more incoming edges. Then they can be removed,as well as their outgoing edges, which are all the edges labeled by 2− and (d + 1)+. Forequilibration requirements, one can also remove all edges labelled 2+ and (d + 1)−.
The procedure going on, one can check that the whole graph will be deleted. Then thereexists no equilibrated cycle. ¤
5.3 End of the proof
We now have the tools necessary to finish the proof of Theorem 5.1.
Proof: (of Th. 5.1). As it has been said before, the direct part is obvious. For the conversepart, from the cluster reduction, it suffices to find a point p removable in Fp according to thepair (Tp, T
′p).
For each line l, we say that a point p is removable in l if p is removable for the pair (Tl, T′l )
of tilings of the unitary d+2-zonotope induced by the pair (T, T ′) from the colorings of l.If T 6= T ′, there exists a line l0 of the diagram associated to T containing a positive point
p0, which can be assumed to be removable in l0. If p0 is removable in Fp0 for (Tp0 , T′p0
), thenwe are done.
Otherwise, there exists a line l1 passing through p0, such that p0 cannot be removed inl1. In this case, there necessarily exists another point p1 which is removable in l1. If p1 isremovable in Fp1 for the pair (Tp1 , T
′p1
), then we are done. Otherwise, we can repeat theprocess.
Assume that there is no point p removable in Fp for the pair (Tp, T′p). With this hypothesis,
the process can be infinitely repeated to construct a sequence (li)i≥0 of lines. Thus there existsa finite subsequence (li)i1≤i≤i2 which is a cycle, i. e.: li1 = li2 .
This cycle is actually a cycle Cline of the line graph. The main idea is the following:
18
(1,1,2,0)
(1,1,1,0)
(0,2,1,1)
(0,2,2,1)
TBB
B ={(0,2,1,1),(0,2,2,1)}
B ={(1,1,1,0),(1,1,2,0)} T’
T’
T
Figure 16: The underlying diagram for two tilings having two incomparable suprema.
From the cycle Cline, one canonically obtains a cycle Cobstacle of the obstacle graph, from themapping which associates to each line l of the diagram, the (coloring given by the) tiling T ′lof the d+2-unitary zonotope. The edge labeling is preserved: if (l, l′) is an edge of Cline, thenthe edge (T ′l , T
′l′) obtained by mapping has the same label as (l, l′).
From the study of the flip graph, the cycle Cobstacle is not equilibrated. One can assumewithout loss of generality, that the label j+ appears in the directed cycle, but label j− doesnot appear. That means that, following the cycle Cline, the jth coordinate changes always inthe same direction. This cannot arise. Thus the process must stop with a line li and a pointpi which can be removed. ¤
5.4 Consequences
5.4.1 Diameter and connectivity
From Theorem 5.1, we obtain a structure of graded poset for the space of tilings. The maxi-mal element is the fully black tiling, the minimal element is the fully white tiling. The rankof a tiling is the number of black points of its diagram. The height of the order is the sum:∑
1≤i1<i2<...<id+1≤d+2 mi1mi2 . . . mid+1. This height above is also the diameter of the space of
tilings.It was known before that for d = 2, or D−d = 1, the space of tilings is a graded poset (See
[9, 3]). As a corollary of Theorem 5.1, we obtain an extension of these results to D − d = 2, i.e. the codimension 2 case.
For D− d = 2, the graded poset given is not a lattice in the general case. See a counterex-ample in Figure 16. This is an important difference with the case D − d = 1.
5.4.2 Uniform random sampling
Consider the Markov process on tilings defined as follows: choose uniformly at random apoint p of the diagram, and a color c (white or black). If the diagram obtained from T givingthe color c to p is the diagram of a tiling T1 then replace T by T1.
Clearly, this process satisfies the hypothesis for ergodicity since the space of tilings isconnected and there exists some loops in the process. Thus the probability distribution pn
obtained after n steps (starting from any distribution, for example, one can take for p0 the
19
Dirac distribution, concentrated on the wholly white tiling) converges to the uniform distri-bution.
6 Perspectives
The decomposition method used here gives an interesting approach for rhombic tilings. Theassociated diagrams give nontrivial results on the spaces of tilings.
The connectivity result presented here looks really encouraging for further studies, andextensions of the flip graph may give more information on the sets of tilings, for exampleflip distances between tilings.
Moreover, the method seems to apply for larger codimensions, or for more general poly-topes.
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